A ray unit for a modulus of a number field is a unit congruent to one modulo every finite prime power in the modulus and positive at every real place in it. The group of ray units is the kernel of the unit group map to the residue and signature group of a modulus.

This degree-four rational map obeys $R(iz)=1/R(z)$ and $R((iz+1)/(1-iz))=e^{2\pi i/3}R(z)$. These pair the generators of the rotational symmetry group of a cube with target rotations, giving combined spatial-isospin symmetry in the Skyrme model. Its spatial half-turn kernel is a Klein four-group. The map gives a cubic four-Skyrmion approximation; preserving only its Wronskian zero set does not guarantee the same rotational symmetry.

The Sullivan model of the connected sum of two complex projective planes has two generators in each of degrees two and three, and none elsewhere. Its rational homotopy groups are therefore exactly the displayed two copies of ${\mathbb{Q}}^2$, with all other positive degrees zero.

independent planar unit steps with uniformly distributed direction have zero vector expected value. The squared Euclidean norm of their sum has expectation $n$, because each diagonal term is one and all cross terms vanish by independence. More generally, zero-mean independent steps with common second moment $s^2$ give mean-square displacement $n{s}^2$; nonzero drift adds a quadratic contribution.

Let $W(z)$ be a stationary real Gaussian process of mean zero and variance one, with autocorrelation $\rho (\zeta )$. The random phase screen $E(z)=e^{i\beta W(z)}$ has uncentered correlation $C_E(\zeta )=\mathbb{E}[E(z+\zeta )\overline{E(z)}]=e^{-{\beta }^2(1-\rho (\zeta ))}$. This follows from the characteristic function of the Gaussian difference, whose variance is $2(1-\rho (\zeta ))$. Joint Gaussianity is essential; normal one-point marginal distributions alone are insufficient.

For the random phase screen $E(z)=e^{i\beta W(z)}$, the Gaussian phase-screen correlation and spectral measure of a stationary random field giveIf $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho }-1$ is integrable, this is $e^{-{\beta }^2}\delta (\nu )$ plus the continuous density $(2\pi {)}^{-1}{e}^{-{\beta }^2}\int (e^{{\beta }^2\rho (\zeta )}-1)e^{-i\nu \zeta }d\zeta$. The total mass is one, split into coherent mass $e^{-{\beta }^2}$ and diffuse mass $1-e^{-{\beta }^2}$. Without the decay and integrability assumptions there may be additional spectral atoms or singular components.

For the random adjacent transposition shuffle on $S_n$ with $n\ge 2$, the position $f(\sigma )={\sigma }^{-1}(i)$ of one label is uniform on $\{1,\dots ,n\}$ in equilibrium, so ${\mathrm{Var}}_{\pi }(f)=(n^2-1)/12$. Each adjacent transposition moves the label with Probability $2/n$, giving $\mathcal{E}(f,f)=(n-1)/n^2$. The Poincare inequality for a reversible Markov chain variational formula therefore gives $\gamma \le 12/[n^2(n+1)]$. In particular the continuous uniform distribution on $(0,1)$ has variance $1/12$, not $1/6$.

For the random adjacent transposition shuffle with $n\ge 2$, the generator $P-I$ is the interchange process on a path graph with each edge rate $1/n$. The Aldous spectral gap theorem reduces its spectral gap to that of the single-label random walk. The eigenfunctions $k\mapsto \cos (\ell \pi (k-1/2)/n)$, for $0\le \ell <n$, have generator eigenvalues $-2(1-\cos (\ell \pi /n))/n$; direct substitution verifies both the interior and endpoint equations. Consequently $\gamma \sim {\pi }^2/n^3$.

An interference graph has one graph vertex for each station and an edge for every pair that cannot transmit simultaneously. Feasible activation schedules are the graph's independent sets.

Carrier-sense multiple access lets a station attempt transmission only when interfering neighbors are inactive. An ideal continuous-time model gives inactive station $r$ an exponential attempt rate $e^{{\theta }_r}$ and unit-rate exponential transmission completions. Detailed balance then gives activation weights $e^{\theta \cdot n}$ on the independent sets of the interference graph.

Throughput is the long-run rate of successful work. In an ideal stationary activation model with unit transmission speed, station $r$ has throughput equal to its stationary active fraction $\mathbb{E}[n_r]$.

For a degree-$n$ totally ramified Galois extension and a uniformizer $\pi$, let $f$ be its minimal polynomial. Dividing the ramification polynomial $f(\pi (1+X))/{\pi }^n$ by its known factor $X$ gives the reduced ramification polynomial. It is monic of degree $n-1$, integral over the extension's valuation ring, and has roots $(\sigma (\pi )-\pi )/\pi$ for nonidentity automorphisms. Its constant coefficient is nonzero in a separable extension. Removing the zero root allows the usual Newton polygon root valuation theorem to apply directly.

The positive-variable system $3x+y=3z$, $x-2y=w$ is partition regular. Its columns for $x,z,w$ sum to zero, and the remaining $y$ column lies in their rational span, satisfying the columns condition. Thus every finite colouring has monochromatic positive $x,y,z,w$ with $2y<x$. A triangular m-p-c set gives the explicit solution $x=3u$, $y=3v$, $z=3u+v$, $w=3u-6v$, with positivity guaranteed by membership in that set.

Suppose real coefficients satisfy $c\le a_n\le C(1+n{)}^m$ for all sufficiently large $n$, where $c,C>0$ and $m\ge 0$. Then the power series $\sum a_n{z}^n$ has radius of convergence one. For $|z|<1$, the comparison test for series and the ratio test dominate it by a polynomial times a geometric sequence. For $|z|\ge 1$, the terms have modulus at least $c$ eventually and violate the term test for divergence. Finite changes to the coefficients do not affect the conclusion.

For a nonzero polynomial $p$, the power series ${\sum }_{n\ge 0}p(n)z^n$ has radius of convergence $1$. If $p$ has degree $d$ and leading coefficient $c\ne 0$, then $p(n)\sim c{n}^d$ and $|p(n+1)/p(n)|\to 1$, so the ratio test applies inside the unit disk. At every point of its boundary, the terms fail to tend to zero. The zero polynomial is the exceptional case with radius of convergence $\infty$.

In the radiogenically heated lava-lake crust model, temperature stays constant precisely when ${\theta }_0=R/B$. Then $h^{\prime }=S(1/h-R)$, and an initially absent crust satisfies the displayed law for $0\le h<1/R$. It follows by integrating $h/(1-Rh)$ from zero to $h$. The crust starts as $\sqrt{2S\tau }$ and tends to $1/R$ only at infinite time. The equilibrium remains thin only if $R\gg 1$.

In the radiogenically heated lava-lake crust model, initially cool liquid warms exponentially and its instantaneous stationary lava-crust thickness $1/(B\theta )$ decreases. The actual crust can cross this moving value, reach a maximum, and melt back to $h_{\ast }=1/R$. Overshoot is not automatic. Linearizing about equilibrium givesIf $B<S{R}^2$ and ${\theta }_0<R/B$, the slowly decaying forced response has positive sign, so the crust eventually approaches equilibrium from above; an initially absent crust must then have overshot. In contrast, initially hot liquid cools, its instantaneous stationary thickness increases, and an initially absent crust cannot cross it from below: at a proposed first crossing its own derivative is zero but the stationary thickness has positive derivative. The hot-start crust therefore grows monotonically. These comparisons assume nonnegative liquid superheat.

In the radiogenically heated lava-lake crust model, the thickness derivative vanishes at $h_e=1/(B\theta )$ when the liquid superheat $\theta$ is positive. This moving curve balances thermal conduction and convective heat flux; it is not itself the evolving crust. Cooling raises the curve and keeps an initially absent crust below it. Warming lowers the curve, allowing lava-crust overshoot under warming. At zero superheat the stationary thickness is infinite, so its finite formula applies only after positive superheat develops.

In a Noetherian ring, every radical of an ideal is an intersection of finitely many prime ideals. A maximal counterexample ideal $J$ cannot be prime. Choose $a,b\notin J$ with $ab\in J$; the strictly larger ideals $J+(a)$ and $J+(b)$ satisfy the result, and $\sqrt{J}=\sqrt{J+(a)}\cap \sqrt{J+(b)}$. This contradicts maximality. The unit ideal uses an empty intersection.

For a constant-luminosity radiative envelope with ideal-gas pressure and Kramers' opacity law, impose power laws on enclosed mass, density, pressure and temperature while retaining $M_r^{\prime }=4\pi r^2\rho$. If their exponents are $m,b,p,d$, respectively, mass conservation gives $m=b+3$, hydrostatic balance gives $d=m-1$, the gas law gives $d=p-b$, and diffusion gives $d-1=2b-2-13d/2$. Solving gives $(m,b,p,d)=(1,-32,-42,-10)/11$. In particular $R/r_c=(T_c/T_R{)}^{11/10}$. Neglecting envelope self-gravity instead sets $M_r$ approximately constant and changes the exponents to $d=-1,b=-13/4,p=-17/4$; that approximation is not an exact solution of mass conservation with nonzero density.